What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (Such deceleration caused one test subject to black out and have temporary blindness.)

Solution:

We are given the following: v_{0}=1000 \ \text{km/h}, v_{f}=0 \ \text{km/h}, \Delta t = 1.1 \ \text{s}.

The acceleration is computed as the change in velocity divided by the change in time.

\begin{align*}
a & = \frac{\Delta v}{\Delta t} \\
a & = \frac{v_{f}-v_{o}}{\Delta t} \\
a & = \frac{\left( 0\ \text{km/h}-1000 \ \text{km/h} \right)\left( \frac{1000 \ \text{m}}{1\ \text{km}} \right) \left( \frac{1\ \text{h}}{3600\ \text{s}} \right)}{1.1\ \text{s}} \\
a & = -252.5\ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

(a) If the rocket sled shown in Figure 4.31 starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is 2100 kg, the thrust T is 2.4 \times 10^{4} N, and the force of friction opposing the motion is known to be 650 N. (b) Why is the acceleration not one-fourth of what it is with all rockets burning?

Solution:

Considering the direction of motion as the positive direction, we are given the following: T=2.4 \times 10^4 \ \text{N}, f=-650 \ \text{N}, and mass, m=2100 \ \text{kg}.

Part A. The magnitude of the acceleration can be computed using Newton’s Second Law of Motion.

\begin{align*}
\Sigma F & =ma \\
2.4\times 10^4 \ \text{N}-650 \ \text{N} & = 2100 \ \text{kg}\times a \\
23350 & = 2100 a \\
\frac{23350}{2100} & = \frac{\cancel{2100} a}{\cancel{2100}} \\
a & = \frac{23350}{2100} \\
a & = 11 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B. The acceleration is not one-fourth of what it was with all rockets burning because the frictional force is still as large as it was with all rockets burning. \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

The same rocket sled drawn in Figure 4.30 is decelerated at a rate of 196 m/s². What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg.

Solution:

Since the rockets are off, the only force acting on the sled is the friction f. This force is against the direction of motion. By using Newton’s Second Law of Motion, we have.

\begin{align*}
\Sigma F & = ma \\
-f & = ma \\
-f & = \left( 2100 \ \text{kg} \right)\left( -196 \ \text{m/s}^{2} \right) \\
-f & = -411600 \ \text{N} \\
f & = 411600 \ \text{N} \\ 
f & = 411.6 \ \text{kN} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The force necessary to produce the given deceleration is 411.6 kN.

In Figure 4.7, the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force F (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force F is removed. How far will the mower go before stopping?

Solution:

We can isolate the mower and expose the forces acting on it. This is the free-body diagram.

There are two forces acting on the mower in the horizontal directions:

1. \textbf{F}:This is the force exerted by the person on the mower, and it is going to the right. This is the first unknown in the problem. We treat this as a positive force since it is directed to the right. The value of this force is F=51 \ \text{N}
2. \textbf{f}: This is the friction force directed opposite the motion of the mower. We treat this as a negative force because it is directed to the left. The value of this force is f=24 \ \text{N}.

Part A. The third force in the figure is the net force, F_{net}. This is the vector sum of the forces F and f. That is

\begin{align*}
F_{net} & =F-f \\
51 \ \text{N} & = F-24 \ \text{N} \\
F & = 51 \ \text{N} +24 \ \text{N} \\
F & = 75 \ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The person should exert a force of 75 N to produce a net force of 51 N.

Part B. When the force F is removed, the friction is now the only force acting on the mower. The friction is acting opposite the direction of motion. The direction of motion is indicated by the blue arrow in the figure below, and the friction force is the red arrow.

Using Newton’s Second Law, we can solve for the deceleration of the mower.

\begin{align*}
\Sigma F & = ma \\
-24 \ \text{N} & = \left( 24 \ \text{kg} \right) \ a \\
\frac{-24 \ \text{N}}{24 \ \text{kg}} & = \frac{\cancel{24 \ \text{kg}}\ \ a}{\cancel{24 \ \text{kg}}} \\
a & = \frac{-24 \ \text{N}}{24 \ \text{kg}} \\
a & = -1 \ \text{m/s}^{2} \\
\end{align*}

Using this deceleration computed above, we can solve for the distance traveled by the mower before coming to stop.

\begin{align*}
\left( v_{f} \right)^{2} & = \left( v_{o} \right)^{2}+2a\Delta x \\
\left( 0 \ \text{m/s} \right)^{2} & = \left( 1.5 \ \text{m/s} \right)^{2}+2\left( -1 \ \text{m/s}^{2} \right)\Delta x \\
 0 & = 2.25-2 \Delta x \\
2 \Delta x & = 2.25 \\
\frac{\cancel{2}\Delta x}{\cancel{2}} & =\frac{2.25}{2} \\
\Delta x & = 1.125 \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}